#include #include #include #include #include using namespace std;
const int maxn = 1e5 + 10;
int kase, t, n;
struct point
};typedef point vector;
typedef vectorpolygon;
vector operator +(vector a, vector b)//
vector operator -(point a, point b)//
vector operator *(vector a, double p)//
vector operator /(vector a, double p)//
bool operator <(const point &a, const point &b)//
const double eps = 1e-4;
int dcmp(double x)//
bool operator ==(const point &a, const point &b)//
double dot(vector a, vector b)//
double length(vector a)//
double cross(vector a, vector b)//
double angle(vector a, vector b)//
double distance(point a, point b)
double distance2(point a, point b)
struct circle
; circle(point c, double r): c(c), r(r) {}
bool operator < (const circle &other) const
bool operator == (const circle &other) const
point point(double a)
bool oncircle(point a) };
circle c[maxn];
circle read_circle()
struct math_line
else
}math_line(circle a, circle b)
math_line(double a = 0, double b = 0, double c = 0): a(a), b(b), c(c) {}
int line_state(const math_line &other) const
else return 1;//相交
} point getlineintersection(const math_line &other) const };
circle center;
int solve()
else if (!(l1.getlineintersection(math_line(c[i], c[i + 1])) == pre)) return -1;
} if (first == 0) return 1;
else return -2; }}
int main(int argc, char const *argv)
return 0;
}
兩個圓的正交圓圓心在直線上，滿足到兩圓心距離之差等於半徑之差。
特殊情況：兩圓重合，只計算乙個即可，兩元同心，必然無解。
那麼兩條不重合直線就可以確定唯一的圓心，去計算r是否相同即可。
你問我所有直線都重合怎麼辦？實際只有一條直線，那麼就是有多解。
對負數開方居然導致presentation error，令我十分吃驚。